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Engineering Exams RRB-JE CS (Chemical Supervisor - Research) Technical Abilities 2025 22.04.25 Paper-2 Shift-1 Medium +1 -0.33
If n12 is the refractive index of medium 1 with respect to medium 2 and n31 is the refractive index of medium 3 with respect to medium 1, then n32, the refractive index of medium 3 with respect to medium 2 will hold the relation .
Correct Answer: A. n32 = n31 × n12
Explanation: To determine the refractive index of medium 3 relative to medium 2 (n32), we look at the relationship between the refractive indices of the media involved. The relative refractive index n32 is defined as the ratio of the absolute refractive index of medium 3 to that of medium 2 (n3/n2). Similarly, n31 represents the ratio of medium 3 to medium 1 (n3/n1), and n12 represents the ratio of medium 1 to medium 2 (n1/n2). By multiplying these two ratios, we get: (n3/n1) × (n1/n2) = n3/n2 Substituting the terms for the relative indices, this simplifies to the mathematical identity n32 = n31 × n12. This shows that the refractive index of a third medium with respect to a second can be calculated by multiplying the refractive index of the third relative to a first with the refractive index of that first medium relative to the second. Thus, option A is the correct relation.
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